Let's break down the problem and solve each part. We're given two sets:

• $A = \{-1, 0, 1\}$
• $B = \{0, 1, 2\}$

We need to analyze the five statements one by one.

1. $n(A \times B) = 6$

• $A \times B$ refers to the Cartesian product of sets $A$ and $B$, which is the set of all ordered pairs where the first element is from $A$ and the second element is from $B$.

  $A \times B = \{(-1, 0), (-1, 1), (-1, 2), (0, 0), (0, 1), (0, 2), (1, 0), (1, 1), (1, 2)\}$

  The total number of elements in $A \times B$ is $|A| \times |B| = 3 \times 3 = 9$. Hence, the statement $n(A \times B) = 6$ is false. The correct number is 9.

2. $(0, 0) \in A \times B$

• The pair $(0, 0)$ consists of elements $0 \in A$ and $0 \in B$. Since both elements exist in their respective sets, $(0, 0) \in A \times B$ is true.

3. $(1, -1) \in A \times B$

• The pair $(1, -1)$ has $1 \in A$, but $-1$ is not an element of $B$. Therefore, $(1, -1) \in A \times B$ is false.

4. $(1, -1) \in B \times A$

• $B \times A$ means the first element is from $B$ and the second element is from $A$.
  • Here, $1 \in B$ but $-1 \in A$, so $(1, -1) \in B \times A$ is true.

5. $\{(0, 1), (1, 2)\} \subseteq A \times B$

• The subset $\{(0, 1), (1, 2)\}$ must be checked against $A \times B$.

  • $(0, 1) \in A \times B$ and $(1, 2) \in A \times B$. Both pairs are in the Cartesian product.

  Therefore, $\{(0, 1), (1, 2)\} \subseteq A \times B$ is true.

Summary of Answers:

1. False
2. True
3. False
4. True
5. True

Would you like to go over any part of this in more detail?

Here are 5 related questions to deepen your understanding:

1. What is the general formula for the number of elements in the Cartesian product of two sets?
2. How would the Cartesian product change if one of the sets were empty?
3. Can an ordered pair in $A \times B$ have elements that are not in either set $A$ or $B$?
4. What is the difference between $A \times B$ and $B \times A$?
5. How can we represent $A \times A$ and what would the size of this product be?

Tip: The Cartesian product is not commutative, meaning $A \times B$ is generally different from $B \times A$.